# Tricky calculus exercise (Now finding a tangent space)

Let $f(x,y,z)=z^3-g(x,y)z+h(x,y)$, with $g,h:\mathbb{R}^2\to \mathbb{R}$ of class $C^1$ and $g(x,y)>0\; \forall (x,y)\in\mathbb{R}^2$. Consider $S=\{(a,b,c)\in\mathbb{R}^3:f(a,b,c)=0\}$, $\Delta =\{(a,b)\in\mathbb{R}^2:4g(a,b)=27h^2(a,b)\}$ and $\pi:S\to\mathbb{R}^2$ defined by $\pi(a,b,c)=(a,b).$

1) Find $\operatorname{Im}(\pi)$.

2) Prove that if $p=(a,b,c)\in S$ then $(a,b)\in \Delta \iff f(p) = \displaystyle\frac{\partial f}{\partial z}(p)=0$.

3) Show that if $p=(a,b,c)\in S$ with $(a,b)\in\mathbb{R}^2\setminus\Delta$ then exists neighborhoods $V$ of $p$ in $\mathbb{R}^2$, $U$ of $(a,b)$ in $\{\mathbb{R}^2-\Delta\}$ and one function $\sigma:U\to V$ of class $C^1$ such that $\pi\circ\sigma|_U = \operatorname{Id}$ in $U$.

4) Consider $g(x,y)=e^{x+y},h(x,y)=\frac{y^2}{x^2+y}\sin(xy)$. Find the space tangent to $(0,1,0)$.

If someone have a solution, please don't post all of it, I just want hint for this one; I will update this post with each solution I find.

Now, I managed to do $(1)$ but I'm having some issues with $(2)$.

Let's start with $(1)$:

$(a,b)\in \operatorname{Im}(\pi)$ if $(a,b,c)\in S$ which means that $c^3-g(a,b)c+h(a,b)=0.$. To have easier notation let $m=g(a,b)$ and $n=h(a,b)$, we can define a new function $s(c)=c^3-mc+n$ with $m>0$, considering the derivatives of this function ($s'(c)=3c^2-m$ and $s''(c)=6c$) seems that this function can return negative and positive values, and by it's continuous (Question: Here I need the continuity of $g,h$?) then by bolzano exists $c$ such that $s(c) =0$. Then for every $(a,b)$ can be found a $c$ such that $f(a,b,c)=0\implies \operatorname{Im}(\pi)=\mathbb{R}^2$.

I didn't finish $(2)$, this is what I have right now:

(a) Suppose that $(a,b)\in \Delta$, then $4g(a,b)=27h^2(a,b)$, which according to the notation I used before menas that $4m-27n^2=0$. Since $p\in S$ then $c^3-mc+n=0$. I need to prove that $\displaystyle\frac{\partial f}{\partial z}(p)=0$ (because since $p\in S$ the condition $f(p)=0$ is trivial), this is, show that $3c^2-m=0$.

I thought that it would be a good idea proceed by substitution, because of $27n^2-4m=0$ I have $n=\displaystyle\frac{2\sqrt{m}}{3\sqrt{3}}$, which means that $c^3-mc+\displaystyle\frac{2\sqrt{m}}{3\sqrt{3}}=0$. Then what?, I don't know how could I get the value of $c$ or $m$ so I can keep proceeding by substitution, and it even seems false because if I take $m=3c^2$ (which I should prove) and plug this value in the former equation I get $-2c^3+\displaystyle\frac{2}{3}|c|$ which doesn't seem necessarily true.

What else could I do here?.

Update 1: Multplying by $c$ the partial derivative $\displaystyle\frac{\partial f}{\partial z}=0$ we have $3c^3-g(a,b)c=0\implies c^3-\frac{1}{3}g(a,b)c=0$ then substracting to $f(a,b,c)=0$ follows $\frac{2}{3}g(a,b)c-h(a,b)=0\implies h^2(a,b)=\frac{4}{9}g(a,b)\implies 27h^2(a,b)=4g(a,b)$.

(I still have no clue how to begin with (3))

Update 2: I believe I know how to solve (3).

The condition $(x,y)\in\mathbb{R}^2\setminus\Delta$ is needed because by (2) we have $\frac{\partial f}{\partial z}\neq 0$, therefore we can define an implicit function $r$ around some open neighborhood $V$ of $(x,y,z)\in S$ (because if $(x,y,z)\in S$ then $f(x,y,z)=0$) such that $r(x,y)=z \iff f(x,y,z)=0$.

Now lets consider $(x,y)\in U$ where $U$ is such that $U\subset V$, defining a function $\sigma (x,y)=(x,y,r(x,y))$ we have that $(x,y,r(x,y))\in S$ then $\pi(x,y,r(x,y))=(x,y)$, this is, $\pi\circ\sigma|_U = \operatorname{Id}$. The uniqueness of $\sigma$ is given by the uniqueness of $r(x,y)$ which is true applying the implicit function theorem. Is this right?.

Now: trying to do (4).

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